Redox Reactions And Electrochemistry Ques 353
Question: In the following reaction, $ 2{{I}^{-}}+Cr_2{{O}^{2-}}_7+14{{H}^{+}}\to I_2+2C{{l}^{3+}}+7H_2O $ Unbalanced parts are
Options:
A) $ {{H}^{+}},,H_2O $
B) $ Cr_2{{O}^{2-}}_7,C{{r}^{3+}} $
C) $ {{I}^{-}},I_2 $
D) None of them are balanced
Show Answer
Answer:
Correct Answer: C
Solution:
[c] (I) $ Cr_2O_7^{2-}+6{{e}^{-}}\to 2C{{r}^{3+}} $ (II) $ 2{{I}^{-}}\to I_2+2{{e}^{-}} $ To balance electron (II) is to be multiplied by (3). Thus, $ \underline{6{{I}^{-}}}+Cr_2{{O}^{2-}}_7+14{{H}^{+}}\to \underline{3I_2}+2C{{r}^{3+}}+7H_2O $ Thus, $ {{I}^{-}} $ and $ I_2 $ are not balanced.
